西安电子科技大学-数字信号处理-试卷C答案

Answer to “Digital Signal Processing of 2005”Problem 1(a) even part: };5.0,1,7,7,5,7,7,1,5.0{---=e X odd part: };5.0,1,3,1,0,1,3,1,5.0{----=o X(b) };20,16,11,94,36,40,31,16,12,0{-----=y (c) MATLAB Programn=-4:2;x=[1 -2 4 6 -5 8 10]; [x11,n11]=sigshift(x,n,2); [x12,n12]=sigshift(x,n,-1); [x13,n13]=sigfold(x,n); [x13,n13]=sigshift(x13,n13,-2); [x12,n12]=sigmult(x,n,x12,n12); [y,n]=sigadd(2*x11,n11,x12,n12); [y,n]=sigadd(y,n,-1*x13,n13)Problem 2(a)w j w j w j w j jw jw e e e e e e X 65424210124)(-----++++++=，()j X e ωis periodic in ω with period 2π(b) MATLAB Program :clear; close all;n = 0:6; x = [4,2,1,0,1,2,4]; w = [0:1:1000]*pi/1000;X = x*exp(-j*n'*w); magX = abs(X); phaX = angle(X); % Magnitude Response Plotsubplot(2,1,1); plot(w/pi,magX);grid;xlabel('frequency in pi units'); ylabel('|X|'); title('Magnitude Response'); % Phase response plotsubplot(2,1,2); plot(w/pi,phaX*180/pi);grid;xlabel('frequency in pi units'); ylabel('Degrees'); title('Phase Response'); axis([0,1,-180,180])(c) Because the given sequence x (n)={4,2,1,0,1,2,4} (n=0,1,2,3,4,5,6) is symmetric about 132N α-==,the phase response ()j H e ω< satisfied the condition ：()3j H e ωαωω<=-=- so the phase response is a linear function in ω.(d) 150,350Hz Hz Ω=-;(e) The difference of amplitude and magnitude response:Firstly, the amplitude response is a real function, and it may be both positive and negative. The magnitude response is always positive.Secondly, the phase response associated with the magnitude response is a discontinuous function. While the associated with the amplitude is a continuous linear function.Problem 3(a) )9.09.01/()1()(211------=z z z z HZero:0 and 1; Pole:-0.6 and 1.5; (b)1116151()212110.61 1.5H z z z--=⨯+⨯+-， 165()((0.6)(1.5))()2121n nh n u n =-+ (c) ROC : 0.6 1.5Z <<，()()()163531215212n nh n u n u n ⎛⎫⎛⎫=⨯--⨯--- ⎪ ⎪⎝⎭⎝⎭ Problem 4(a) y(n)={50,44,34,52};(b) y(n)={5,16,34,52,45,28,0}; (c) N=6;(d) MATLAB Program :Function y=circonv(x1,x2,N) If (length(x1)>N)error(“N must not be smaller than the length of sequence ”) elsex1=[x1,zeros(1,N-length(x1))]; endif(length(x2)>N)error(“N must not be smaller than the length of sequence ”)elsex2=[x2,zeros(1,N-length(x2))]; endy1=dft(x1,N).*dft(x2,N); y=idft(y,N);(e) DTFT is discrete in time domain, but continuous in frequency domain. The DFT is discrete both in time and frequency domain.The FFT is a very efficient method for calculating DFT.Problem 5(a) Direct form II uses the little delay and it can decrease the space of the compute. (b)The advantage of the linear-phase form:1. For frequency-selective filters, linear-phase structure is generally desirable to have a phase-responsethat is a linear function of frequency.2. This structure requires 50% fewer multiplications than the direct form. (c) Block diagrams are shown as under:1z -1z -1z -1z -1z -1z -()x n )n()x n1-1-Problem 6(a) we use Hamming or Blackman window to design the bandpass filter because it can provide us attenuationexceed 60dB .(b) According to Blackman window :first, Determine transition width =p s W W - ;second, Determine the type of the window according to s A ;third, Compute M using the formula MW W p s π2=- ; fourth, Compute ideal LPF2sp W W Wc +=;fifth, design the window needed, multiply point by point; sixth, determines p A R ,(c) MATLAB Program :%% Specifications about Blackman window: ws1 = 0.2*pi; % lower stopband edge wp1 = 0.3*pi; % lower passband edge wp2 = 0.6*pi; % upper passband edge ws2 = 0.7*pi; % upper stopband edge Rp = 0.5; % passband ripple As = 60; % stopband attenuation %tr_width = min((wp1-ws1),(ws2-wp2));M = ceil(6.6*pi/tr_width); M = 2*floor(M/2)+1, % choose odd M n = 0:M-1;w_ham = (hamming(M))';wc1 = (ws1+wp1)/2; wc2 = (ws2+wp2)/2; hd = ideal_lp(wc2,M)-ideal_lp(wc1,M); h = hd .* w_ham;[db,mag,pha,grd,w] = freqz_m(h,1); delta_w = pi/500;Asd = floor(-max(db([1:floor(ws1/delta_w)+1]))), % Actual AttnRpd = -min(db(ceil(wp1/delta_w)+1:floor(wp2/delta_w)+1)), % Actual passband ripple （5） %%% Filter Response Plotssubplot(2,2,1); stem(n,hd); title('Ideal Impulse Response: Bandpass'); axis([-1,M,min(hd)-0.1,max(hd)+0.1]); xlabel('n'); ylabel('hd(n)') set(gca,'XTickMode','manual','XTick',[0;M-1],'fontsize',10) subplot(2,2,2); stem(n,w_ham); title('Hamming Window'); axis([-1,M,-0.1,1.1]); xlabel('n'); ylabel('w_ham(n)')set(gca,'XTickMode','manual','XTick',[0;M-1],'fontsize',10) set(gca,'YTickMode','manual','YTick',[0;1],'fontsize',10)subplot(2,2,3); stem(n,h); title('Actual Impulse Response: Bandpass'); axis([-1,M,min(hd)-0.1,max(hd)+0.1]); xlabel('n'); ylabel('h(n)') set(gca,'XTickMode','manual','XTick',[0;M-1],'fontsize',10)subplot(2,2,4); plot(w/pi,db); title('Magnitude Response in dB');axis([0,1,-As-30,5]); xlabel('frequency in pi units'); ylabel('Decibels') set(gca,'XTickMode','manual','XTick',[0;0.3;0.4;0.5;0.6;1])set(gca,'XTickLabelMode','manual','XTickLabels',['0';'0.3';'0.4';'0.5';'0.6';'1'],... 'fontsize',10)set(gca,'TickMode','manual','YTick',[-50;0])set(gca,'YTickLabelMode','manual','YTickLabels',['-50';'0']);gridProblem 7Firstly, we use the given specifications ofs p s p A R ,,,ωω to design an analog lowpass IIR filter.Secondly, we change the analog lowpass IIR filter into the analog highpass IIR filter. Thirdly, we change the analog highpass IIR filter into the digital highpass IIR filter.。